Question

It$X$is a Poisson random variable with a mean$100$, then$PX>120$is approximately

$\left(a\right).02,$

$\left(b\right).5$or

$\left(c\right).3?$

Short answer

Then,$PX>120$is approximately $\left(a\right)\mathbf{.}\mathbf{02}.$

Step-by-step solution

Given Information.

$X$is a Poisson random variable with a mean$100$.

Explanation.

Suppose that$X_1,\dots ,X_{100}$is independent and equally distributed with the distribution$Pois\left(1\right)$. Therefore

$X=\sum _i{X}_i~Pois\left(100\right)$

Using the Central Limit Theorem, we can approximate that

$X~AN\left(nE\left(X_1\right),Var\left(X_1\right)\right)=AN(100,1)$

Therefore, we have an approximation

$P(X>120)=P\left(\frac{X-nE\left(X_1\right)}{\sqrt{Var\left(X_1\right)}}>\frac{120-nE\left(X_1\right)}{\sqrt{Var\left(X_1\right)}}\right)$

$=P\left(\frac{X-nE\left(X_1\right)}{\sqrt{Var\left(X_1\right)}}>20\right)\approx 1-\Phi \left(20\right)=0.02$

so the right answer is$\left(a\right)$.